The equation of the plane through the points `(1,1,1),(1,-1,1) and (-7,-3, -5)` is parallel to the axis of y i.e. perpendicular to xz-plane.

Answer

To find the equation of the plane through the points \( (1, 1, 1) \), \( (1, -1, 1) \), and \( (-7, -3, -5) \) that is parallel to the y-axis (and hence perpendicular to the xz-plane), we can follow these steps:

### Step 1: Identify the Points
The three points given are:
- \( P_1(1, 1, 1) \)
- \( P_2(1, -1, 1) \)
- \( P_3(-7, -3, -5) \)

### Step 2: Set Up the Equation of the Plane
The general equation of a plane can be expressed in the form:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
Where \( (x_1, y_1, z_1) \) is a point on the plane, and \( (a, b, c) \) is the normal vector to the plane.

### Step 3: Find Vectors in the Plane
We can find two vectors in the plane by subtracting the coordinates of the points:
1. Vector \( \vec{v_1} = P_2 - P_1 = (1 - 1, -1 - 1, 1 - 1) = (0, -2, 0) \)
2. Vector \( \vec{v_2} = P_3 - P_1 = (-7 - 1, -3 - 1, -5 - 1) = (-8, -4, -6) \)

### Step 4: Find the Normal Vector
The normal vector \( \vec{n} \) to the plane can be found using the cross product of the two vectors \( \vec{v_1} \) and \( \vec{v_2} \):
\[ \vec{n} = \vec{v_1} \times \vec{v_2} \]
Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 0 \\ -8 & -4 & -6 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{n} = \hat{i}((-2)(-6) - (0)(-4)) - \hat{j}((0)(-6) - (0)(-8)) + \hat{k}((0)(-4) - (-2)(-8)) \]
\[ = \hat{i}(12) - \hat{j}(0) + \hat{k}(-16) \]
Thus, the normal vector is:
\[ \vec{n} = (12, 0, -16) \]

### Step 5: Write the Equation of the Plane
Using point \( P_1(1, 1, 1) \) and the normal vector \( (12, 0, -16) \):
\[ 12(x - 1) + 0(y - 1) - 16(z - 1) = 0 \]
Simplifying this gives:
\[ 12x - 12 - 16z + 16 = 0 \]
\[ 12x - 16z + 4 = 0 \]

### Step 6: Simplify the Equation
Dividing the entire equation by 4:
\[ 3x - 4z + 1 = 0 \]

### Final Equation
Thus, the equation of the plane is:
\[ 3x - 4z + 1 = 0 \]